LeetCode(四)链表206. Reverse Linked List&2. Add Two Numbers

最新推荐文章于 2025-12-06 07:43:37 发布

转载最新推荐文章于 2025-12-06 07:43:37 发布 · 220 阅读

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本文介绍两种常见的链表操作技巧：链表的翻转与两个数相加。通过具体实例展示了如何实现链表的翻转，并给出了两数相加的具体算法实现，包括进位处理等细节。

206. Reverse Linked List

翻转一个链表
解法如下

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode reverseList(ListNode head) {
        ListNode newhead=null;
        while(head!=null){
            ListNode next=head.next;
            head.next=newhead;
            newhead=head;
            head=next;
        }
        return newhead;
    }
}

2. Add Two Numbers

题目要求

You are given two non-empty linked lists representing two non-negative
integers. The digits are stored in reverse order and each of their
nodes contain a single digit. Add the two numbers and return it as a
linked list.

You may assume the two numbers do not contain any leading zero, except the number 0 itself.

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4) Output: 7 -> 0 -> 8

解法如下

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        ListNode head1=l1;
        ListNode head2=l2;
        ListNode res=new ListNode(0);
        int sum=0;
        ListNode d=res;
        while(head1!=null||head2!=null){
            sum/=10;
            if(head1!=null){
                sum+=head1.val;
                head1=head1.next;
            }
            if(head2!=null){
                sum+=head2.val;
                head2=head2.next;
            }
            d.next=new ListNode(sum%10);
            d=d.next;
        }
        if(sum/10==1){//进位
            d.next=new ListNode(1);

        }
        return res.next;
    }
}